Q.The equation of the line joining the point $(3,5)$ to the point of intersection of the lines $4x+y-1=0$ and $7x-3y-35=0$ is equidistant from the points $(0,0)$ and $(8,34)$.

Concept: Distance from a point to a line.

Step 1 – Find intersection point

Solve $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$.

From the first, $y = 1 - 4x$. Substitute into the second:

$7x - 3(1 - 4x) - 35 = 0 \implies 7x - 3 + 12x - 35 = 0 \implies 19x = 38 \implies x = 2$.

Then $y = 1 - 8 = -7$. So intersection is $(2, -7)$.

Step 2 – Equation of the required line

Line through $(3,5)$ and $(2,-7)$: slope $m = \frac{-7 - 5}{2 - 3} = \frac{-12}{-1} = 12$.

Equation: $y - 5 = 12(x - 3) \implies y = 12x - 31$, or $12x - y - 31 = 0$.

Step 3 – Equidistant condition …

Concept and Intuition

The distance from a point to a line is the perpendicular distance. When a line is equidistant from two points, it means the perpendicular distances from those points to the line are equal. This often happens in two situations: either the line is parallel to the segment joining the two points and lies midway between them, or the line actually passes through the midpoint of the two points. But here, the line is not free — it must pass through a fixed point $(3,5)$ and also through the intersection of two given lines. So we first find that intersection, then write the family of lines through it and $(3,5)$, and impose the equidistance condition.

Step-by-step solution

1. Find the intersection point of the two given lines

We solve:

From the first equation: $y = 1 - 4x$.

Substitute into the second:

Then $y = 1 - 4(2) = 1 - 8 = -7$.

So the intersection point is $(2, -7)$.

2. Equation of the line joining $(3,5)$ and $(2,-7)$

The slope is:

Using point-slope form with $(3,5)$:

So the line is $12x - y - 31 = 0$.

3. Condition of equidistance from $(0,0)$ and $(8,34)$ …